Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a two-dimensional surface in $\mathbb{R}^3$, I thought that the total mean curvature was equal to the first variation of area:

$$\frac{d}{dt}SA(t)\Big\vert_{t\to 0} = \int H dA,$$

where $SA(t)$ is the surface area of the surface after flowing it along its normal vector field for time $t$.

But when I try this formula for a cylinder of unit height and radius $r$, I get that $SA(t) = 2\pi (r+t)$, $H=\frac{1}{2r}$, and

$$2\pi \stackrel{?}{=} 2\pi r \frac{1}{2r} = \pi.$$

Where have I gone wrong? Am I missing a factor of two in the first variation of area formula?

share|improve this question
If $n$ is an unit normal field to the surface $S$ then, for any $\phi\in C_c^\infty(S),$ you should get $\left.\dfrac{d}{dt}\right|_{t=0}\textrm{Surface}(\{x+t\phi(x) n(x):x\in S\})=\mathbf{2}\int_S\phi(x)H(x)d\sigma(x).$ –  Giuseppe Tortorella Jun 12 '12 at 7:25
Ah! You're right, it looks like the paper I read used the annoying alternate definition of "mean" curvature as $H=k_1+k_2$. Thanks! –  user7530 Jun 12 '12 at 7:44
You could write that up as an answer and accept it so the question doesn't remain unanswered. –  joriki Jun 12 '12 at 9:49

1 Answer 1

up vote 0 down vote accepted

As Giusepppe mentions in the comments, the right hand side in the formula is indeed incorrect. It should be

$$\int 2 H dA.$$

Sometimes in the literature mean curvature is defined as $H = k_1 + k_2$ instead of $(k_1+k_2)/2$ and this discrepancy was the source of my error.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.